Respuesta :
Answer: 15.2 grams
Explanation:
The balanced chemical equation is :
[tex]FeCl_3(aq)+3AgNO_3(aq)\rightarrow 3AgCl(s)+Fe(NO_3)_3(aq)[/tex]
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} FeCl_3=\frac{32.4g}{162.2g/mol}=0.200moles[/tex]
[tex]\text{Moles of} AgNO_3=\frac{18.0g}{169.87g/mol}=0.106moles[/tex]
According to stoichiometry :
3 moles of [tex]AgNO_3[/tex] require = 1 mole of [tex]FeCl_3[/tex]
Thus 0.106 moles of [tex]AgNO_3[/tex] will require=[tex]\frac{1}{3}\times 0.106=0.0353moles[/tex] of [tex]FeCl_3[/tex]
Thus [tex]AgNO_3[/tex] is the limiting reagent as it limits the formation of product and [tex]FeCl_3[/tex] is the excess reagent.
As 3 moles of [tex]AgNO_3[/tex] give = 3 moles of [tex]AgCl[/tex]
Thus 0.106 moles of [tex]AgNO_3[/tex] give =[tex]\frac{3}{3}\times 0.106=0.106moles[/tex] of [tex]AgCl[/tex]
Mass of [tex]AgCl=moles\times {\text {Molar mass}}=0.106moles\times 143.32g/mol=15.2g[/tex]
Thus 15.2 g of [tex]AgCl[/tex] will be produced from the given masses of both reactants.
